Answer:
From \[\tau =I\alpha \] or \[\alpha =\frac{\tau }{I}\]
As \[\tau =cons\tan t\], \[\alpha \propto
\frac{1}{I}\] \[\therefore \] \[\frac{{{\alpha
}_{c}}}{{{\alpha }_{s}}}=\frac{{{I}_{s}}}{{{I}_{c}}}\]
Now moment of inertia of cylinder about its axis is \[{{I}_{c}}=\frac{1}{2}M{{R}^{2}}\]
and moment of inertia of hollow sphere about its diameter
is \[{{I}_{s}}=\frac{2}{3}M{{R}^{2}}\]
\[\therefore \] \[\frac{{{\alpha
}_{c}}}{{{\alpha }_{s}}}=\frac{2/3\,M{{R}^{2}}}{1/2M{{R}^{2}}}=\frac{4}{3}\] \[\therefore
\] \[{{\alpha }_{c}}>{{\alpha }_{s}}\]
Hence, the cylinder will acquire greater speed compared to
the hollow sphere.
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